Math  /  Algebra

QuestionBUSINESS A Mason jar company wants to increase the volume of its cylindrical jars by 6 cubic inches. The company's designer wants a formula that states the height hh of a jar given its volume VV and radius rr. The volume of a cylindrical jar is modeled by the equation V=πr2hV=\pi r^{2} h. Part A What formula should be used to find the height hh ? A. h=πr2h=\pi \cdot r^{2} B. h=Vπrh=\sqrt{\frac{V}{\pi r}} C. h=Vπr2h=\frac{V}{\pi r^{2}} D. h=Vr2h=\frac{V}{r^{2}}
Part B If the radius of the jar is 1.5 inches and the original volume is 30 cubic inches, then what height should the company make its new jar to increase the volume by 6 cubic inches?

Studdy Solution

STEP 1

1. The volume of a cylindrical jar is given by the formula V=πr2h V = \pi r^2 h .
2. We need to find a formula for the height h h in terms of V V and r r .
3. The original volume of the jar is 30 cubic inches, and it needs to be increased by 6 cubic inches.
4. The radius of the jar is 1.5 inches.

STEP 2

1. Derive the formula for height h h from the given volume formula.
2. Calculate the new height for the increased volume.

STEP 3

Start with the volume formula for a cylinder: V=πr2h V = \pi r^2 h

STEP 4

Solve for h h by dividing both sides by πr2 \pi r^2 : h=Vπr2 h = \frac{V}{\pi r^2}

STEP 5

The correct formula for height h h is: h=Vπr2 \boxed{h = \frac{V}{\pi r^2}}

STEP 6

For Part B, calculate the new volume: Original volume V=30 V = 30 cubic inches. Increase in volume = 6 cubic inches. New volume Vnew=30+6=36 V_{\text{new}} = 30 + 6 = 36 cubic inches.

STEP 7

Use the formula h=Vπr2 h = \frac{V}{\pi r^2} to find the new height with r=1.5 r = 1.5 inches and Vnew=36 V_{\text{new}} = 36 cubic inches: h=36π(1.5)2 h = \frac{36}{\pi (1.5)^2}

STEP 8

Calculate (1.5)2=2.25 (1.5)^2 = 2.25 .

STEP 9

Substitute back into the formula: h=36π×2.25 h = \frac{36}{\pi \times 2.25}

STEP 10

Calculate the height: h=362.25π367.06865.09 h = \frac{36}{2.25\pi} \approx \frac{36}{7.0686} \approx 5.09
The new height should be approximately: 5.09 inches \boxed{5.09 \text{ inches}}

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